
In order to be differentiable everywhere,
must first be continuous everywhere, which means the limits from either side as
must be the same and equal to
. By definition,
, and


so we need to have
.
For
to be differentiable at
, the derivative needs to be continuous at
, i.e.

We then need to have

Then

Answer:
S/2πr-r = h
Step-by-step explanation:
S = 2πr(h + r)
S/2πr = h+r
S/2πr - r = h
Answer: Assuming the plumber will round up the hours, the answer is 335.
Step-by-step explanation:
You can use this equasion to solve the problem: 75+52x=c
In this case, x would equal the number of hours he would work and c would represent the final cost.
Basically, you can just multiply the hourly cost by the hours he works and then add the service fee. Hope this helped!
If u do the formula m=y^2-y^1/x^2-x^1 then u have 3-2/6-0 and that simplifies to 1/6 and that is your answer
answer: 1/6
A = 57
45 + 12 = 57
57 - (-12) = 45